Answer:
222.5 Gpa
Explanation:
From definition of engineering stress,
where F is applied force and A is original area
Also, engineering strain, where l is original area and
is elongation
We also know that Hooke's law states that
Since A=20 mm* 20 mm= 0.02 m*0.02 m
F= 89000 N
l= 100 mm= 0.1 m
By substitution we obtain
Answer:
a) ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
Answer:
BOD5 = 200 mg/L
Explanation:
given data
diluted = 1% = 0.01
time = 5 day
oxygen consumption = 2.00 mg · L−1
solution
we get here BOD5 that is BOD after 5 day
and here total volume is 100% = 1
so dilution factor is = 100
so BOD5 is
BOD5 = oxygen consumption × dilution factor
BOD5 = 2 × 100
BOD5 = 200 mg/L
